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Title: JEE MAINS SHIFT @ FEBRUARY
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FIITJEE
Solutions to JEE(Main) -2023
Test Date: 1st February 2023 (Second Shift)

PHYSICS, CHEMISTRY & MATHEMATICS
Paper - 1
Time Allotted: 3 Hours


Maximum Marks: 300

Please read the instructions carefully
...


Important Instructions:
1
...


2
...
Each subject (PCM) has 30 questions
...


3
...
Part-A is Physics, Part-B is Chemistry and
Part-C is Mathematics
...


4
...


5
...


6
...
Each question carries +4 marks for correct answer and –1 mark for wrong
answer
...


Section-B (1 – 10) contains 10 Numerical based questions
...
Each question carries +4 marks for correct
answer and –1 mark for wrong answer
...
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...
com
...
Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


Given below are two statements : One is labelled as Assertion A and the other is labelled as
Reason R
...

Reason R : Voltmeter with higher resistance will draw smaller current than voltmeter with lower
resistance
...

(A) Both A and R are correct and R is the correct explanation of A
(B) A is not correct but R is correct
(C) Both A and R are correct but R is not the correct explanation of A
(D) A is correct but R is not correct

Q2
...
The distance between images formed by the mirror is
_________
...


A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of
magnetic field
...
By changing the magnitude of the magnetic field within the coil
...
By changing the area of coil within the magnetic field
...
By changing the angle between the direction of magnetic field and the plane of the coil
...
By reversing the magnetic field direction abruptly without changing its magnitude
...


The escape velocities of two planets A and B are in the ratio 1 : 2
...


If the velocity of light c, universal gravitational constant G and Planck’s constant h are chosen as
fundamental quantities
...
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...


The Young’s modulus of a steel wire of length 6m and cross-sectional area 3 mm 2, is 2 × 1011
2
N/m
...
A block of mass 4kg is attached
1
to the free end of the wire
...
The elongation of wire is (Take g on the earth = 10m/s ) :
(A) 1 cm
(B) 0
...
1 mm

Q7
...

(B) It works as a voltage regulation in both forward and reverse bais
...

(D) It works as a voltage regulator only in forwards bias
...


Given below are statements : One is labelled as Assertion A and the other is labelled as Reason
R
...
One of them is hollow and
another is solid, and both have the same radii
...

Reason R : Capacitance of metallic sphere depend on the radii of sphere
...

(A) A is false but R is true
(B) A is true but R is false
(C) Both A and R are true but R is not the correct explanation of A
(D) Both A and R are true and R is the correct explanation of A

Q9
...
Then, in second case, the same modulating signal is
superimposed with different carrier signal of amplitude 2Y V
...


Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire
of resistance R would be :
n  1 R
n  1 R
(A)
(B)
n2
n
2
n  1 R
n R
(C)
(D)
n 1
 2n  1

Q11
...
The energy released in this process, will be :
(Given Rch = 13
...
4 eV
(B) 40
...
5 eV
(D) 13
...
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...


As shown in the figure, a long straight conductor with

semicircular arc of radius m is carrying current I = 3A
...
at the centre O of the
arc is :
(The permeability of the vacuum = 4  10 7 NA 2 )
(A) 3 T
(B) 6 T
(C) 4 T
(D) 1 T

O

I = 3A

I = 3A

Q13
...


For three low density gases A, B, C pressure
versus temperature graphs are plotted while
keeping them at constant volume, as shown
in the figure
...


Q16
...
When the light of frequency 2f 0 is incident on the metal
plate, the maximum velocity of photoelectrons is 1
...
The ratio of 1 to 2 is :
(A)

1 1

2 2

(B)

1 1

2 4

(C)

1
1

2 16

(D)

1 1

2 8

Figures (a), (b), (c) and (d) show variation of force with time
...
5

0
...
0

t(s)

0

2
...
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...
5

0
...
0

The impulse is highest in figure
...


2
...

T

2

T

(A)

2

(B)

O
T

O

L

2

T

(C)

L

2

(D)

O

Q18
...
For  s  0
...
3 N
(B) 25
...
7 N
(D) 20 N

F
30 

1
Q19
...
When the temperature of cold
3
1
reservoir raised by x, its efficiency decreases to
...
if the temperature hot reservoir
6
is 99 C, will be :
(A) 66 K
(B) 16
...


For a body projected at an angle with the horizontal from the ground, choose the correct
statement
...

(B) The Kinetic Energy (K
...
) is zero at the highest point of projectile motion
...

(D) The horizontal component of velocity is zero at the highest point
...
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...
The answer to each question is rounded off to the
nearest integer value
...


In the given circuit, the value of

10V

I1  I3
is ________
I2

10 

20V

I1

10 
I2
I3

10 

MR2

...
The value of x is _________
...


Moment of inertia of a disc of mass M and radius ‘R’ about any of its diameter is

Q3
...
4 wbm -2,
about an axis which is parallel to one of the side of the coil and perpendicular to the direction of
field
...
(Take  
)
7

Q4
...
The speed of water coming out through the tap of cross section area 500
dh
mm2 is 30cm/s
...
The value of x will be _______
...


A block is fastened to a horizontal spring
...
The energy of the block at x =
-1
5cm is 0
...
The spring constant of the spring is ________Nm
...


Nucleus A having Z = 17 and equal number of protons and neutrons has 1
...

Another nucleus B of Z = 12 has total 26 nucleons and 1
...

The difference of binding energy of B and A will be _______ MeV
...


A cubical volume is bounded by the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a
...
Where E0  4  104 NC1 m1
...
The value of Q is ________
...


(Take 0  9  1012 C2 / Nm2 )
For a train engine moving with speed of 20 ms-1, the driver must apply brakes at a distance of
500m before the station for the train to come to rest at the station
...
The value of x is
_______
...
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...


A force F  (5  3y 2 ) acts on a particle in the y-direction, where F is in newton and y is in meter
...


Q10
...
2 is inserted
infront of slit S1
...
Due to the
insertion of the plate, central maxima is shifted by a distance of
x 0
...
The
value of the x is ________
...
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...
Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


Given below are two statements: one is labelled as Assertion (A) and the other is labelled as
Reason (R)
Assertion (A): Cu2+ in water is more stable than Cu+
2+
+
Reason (R): Enthalpy of hydration for Cu is much less than that of Cu
In the light of the above statements, choose the correct answer from the options given below:
(A) Both (A) and (R) are correct but (R) is not the correct explanation of (A)
(B) (A) is correct but (R) is not correct
(C) Both (A) and (R) are correct and (R) is the correct explanation of (A)
(D) (A) is not correct but (R) is correct

Q2
...

Reason (R): Amino acids exist in zwitter ion form in aqueous medium
...


The correct order of bond enthalpy (kJ mol1) is:
(A) Si –Si > C – C > Ge – Ge > Sn – Sn
(B) C – C > Si – Si > Ge – Ge > Sn – Sn
(C) C – C > Si – Si> Sn – Sn > Ge – Ge
(D) Si – Si > C – C > Sn – Sn > Ge – Ge

Q4
...

O
H

NaHSO3, dil
...
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H , C =
...

'X' is

Q7
...

Statement II: Sulphanilic acid gives red colour in Lassigne’s test for extra element detection
...
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In a reaction
OH

OH

COOCH3

OCOCH3
COOH

'Y'

COOH
'X'

Reagents ‘X’ and ‘Y’ respectively are:
(A) (CH3CO)2O/H+ and CH3OH/H+,
(B) CH3OH/H+,  and (CH3CO)2O/H+
(C) CH3OH/H+, and CH3OH/H+,
+
+
(D) (CH3CO)2O/H and (CH3CO)2O/H
Q9
...


Which element is not present in Nessler’s reagent?
(A) Oxygen
(B) Mercury
(C) Iodine
(D) Potassium

Q11
...


The graph which represents the following reaction is:


OH
  C6H5 3 C  OH
 C6H5 3 C  Cl 
Pyridine

rate

rate

(A)

(B)

[(C6H5)3C-Cl]

[OH]

rate

rate

(C)

(D)

[(C6H5)3C-Cl]

[Pyridine]

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-11

Q13
...
Most stable of them is:
HO
OH
O

HO
(A)

O

O

HO
H

H
O

HO

OH

HO
O

Q15
...


O

HO

HO
(C)

O

(B)

OH

H

OH

O

OH

The Industrial activity held least responsible for global warming is:
(A) Electricity generation in thermal power plants
(B) Steel manufacturing
(C) Industrial production of urea
(D) manufacturing of cement
In figure, a straight line is given for Freundrich Adsorption
1
(y = 3x +2
...
The vale of and log K are respectively
...
3 and 0
...
505
(C) 0
...
505
(D) 3 and 0
...


The complex cation which has two isomers is :
(A) [Co(NH3)5Cl]2+
(C) [Co(NH3)5NO2]2+

Q17
...


The effect of addition of helium gas to the following reaction in equilibrium state, is
PCl5 (g)  PCl3 (g)  Cl2 (g)
(A) The equilibrium will shift in the forward direction and more of Cl 2 and PCl3 gases will
be produced
...

(C) Helium will deactivate PCl5 and reaction will stop
...


Q19
...

Reason (R): On aging, KOH solution absorbs atmospheric CO2
...
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...


O-O bond length in H2O2 is X than the O-O bond length in F2O2
...

Choose the correct option for X and Y from those given below:
(A) X – shorter, Y = longer
(B) X- longer, Y – shorter
(C) X- shorter, Y- shorter
(D) X- longer, Y- longer

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-13

SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions
...

Q1
...
(Nearest integer)
[Given: molar mass of Br2 = 160 g mol1
Atomic mass of C = 12 g mol1, Atomic mass of Cl  35
...
2g cm3
Density of CCl4 = 1
...


0
...
The temperature of
calorimeter system (including the water) is found to rise by 0
...
The heat evolved during
combustion of ethane at constant pressure is_______ kJ mol1
...
3 JK1 mol1
...

Atomic mass of C and H are 12 and 1g mol1 respectively]

Q3
...
Half life of this reaction is 50 min
...

(Nearest integer)

Q4
...

The depression in freezing point of such water is___________ 103 oC
...
m
...

[Given: Molal depression constant and density of water are 1
...


Q5
...

(A) CuCO3
(B) Cu2S
(C) Cu2O
(D) FeO

Q6
...

(A) Chloroliazepoxide
(B) Veronal
(C) Valium
(D) Salvarsan

Q7
...

OH
CH3
CH3

O

Testosterone
The total number of asymmetric carbon atom/s in testosterone is___________
...
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The conductivity of this solution at
298 K is__________108 Sm1
...
91013 at 298 K
 0Ag  6  103 Sm2 mol1

Q8
...


The spin only magnetic moment of [Mn(H2O)6]2+ complexes is_______B
...
of Mn is 25)

Q10
...
0 and 2
...
The ratio of densities of lattices fcc to bcc for
the metal M is___________
...
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Each question has four choices (A), (B), (C) and
(D), out of which ONLY ONE option is correct
...


The sum of the absolute maximum and minimum
f  x   x 2  5x  6  3x  2 in the interval  1, 3  is equal to :
(A) 10
(C) 13

Q2
...


values

 x, y  : xy  8, 1  y  x  is :
2

13
3
7
(D) 16loge 2 
3

(B) 8loge 2 

 1 
Let f : R  0,1  R be a function such that f  x   f 
  1  x
...


The number of integral values of k, for which one root of the equation 2x 2  8x  k  0 lies in the
interval 1, 2  and its other root lies in the interval  2, 3  , is :
(A) 0
(B) 1
(C) 2
(D) 3

Q5
...


(B) 4loge 2  2
2

(D) 4  loge 2   2



ˆ b  ˆi  kˆ and c  ˆi  2ˆj  3kˆ be three given vectors
...
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

...

2
1
(B) n  S   1 and the elements in S is more than
...

2

Let P  S  denote the power set of S  1, 2, 3,
...
Define the relations R1 and R2 on P  S  as

Q8
...
Then :
(A) both R1 and R2 are not equivalence relations
(B) only R1 is an equivalence relation
(C) both R1 and R2 are equivalence relations
(D) only R2 is an equivalence relation
Q9
...
 x7 be in an A
...
with common difference d
...
,x 7 is 4 and the mean is x , then x  x 6 is equal to :

1 
(B) 18 1 

3

(D) 34


8 
(A) 2  9 

7

(C) 25

Q10
...
Then

2  y 0  x 0  is equal to :

(A) 9
(C) 3
Q11
...


The sum

11e 7

2
2e
13e 5
(C)

4
4e

11e 7

4
2
2e
13e 5
(D)

4
4
4e

(A)

Q13
...
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...


Two dice are thrown independently
...
Then :
(A) the number of favourable cases of the events A, B and C are 15, 6 and 6 respectively
(B) the number of favourable cases of the event  A  B   C is 6
(C) A and B are mutually exclusive
(D) B and C are independent

Q15
...
If the complex number
and a  ib lies on the circle z  1  2z , then a possible value of
integer function, is :
1
(A)
2
1
(C) 
2

, where  t  is greatest

(D) 1


4 dx is :
The value of the integral 
 2  cos 2x


(A)



x

4

2

(B)

6 3

2
3 3

2

(C)

Q17
...


1  a 

1  ai
is of unit modulus
bi


6

(D)



2
12 3



Let x  exp x y  be the solution of the differential equation





2x 2 y dy  1  xy2 dx  0, x  0, y  2   loge 2
...


(B) 3
(D) 0

For the system of linear equations x  y  z  1, x  y  z  1, x  y  z   , which one of the
following statements is NOT correct?
(A) It has infinitely many solutions if   2 and   1
(B) It has no solution if   2 and   1
3
if   2 and   1
4
(D) It has infinitely many solutions if   1 and   1

(C) x  y  z 

Q19
...



b  ˆi  3ˆj  5kˆ be two vectors
...
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35


17
(C) Projection of a on b is
and the direction of the projection vector is opposite to the
35

direction of b
...

35

Q20
...
If d is the distance of P from the point

 7, 1, 1 , then
250
83
25
(C)
83

(A)

d2 is equal to :

15
53
250
(D)
82

(B)

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-19

SECTION - B
(Numerical Answer Type)
This section contains 10 Numerical based questions
...



Q1
...




1  5cos x

k
, then k is equal to………
16

Let x  y  yz  1 be the equation of a plane passing through the point

 3,  2, 5 

and

perpendicular to the line joining the points 1, 2, 3  and  2, 3, 5 
...

Q3
...


Q4
...


The point of intersection C of the plane 8x  y  2z  0 and the line joining the points A  3,  6, 1
B  2, 4,  3  divides

and



the

line

segment

AB

internally

in

the

ratio

k : 1
...


The sum of the common terms of the following three arithmetic progressions
3,7,11,15,
...
,359 and
2,7,12,17,
...


If the x-intercept of a focal chord of the parabola y 2  8x  4y  4 is 3, then the length of this chord
is equal to……
...


powers of 2  2 , be 21
...
P
...

x  2 log 3

Q9
...
If
a2 b2



the tangent to E at the point P in the first quadrant passes through the point 0, 4 3



2

intersects the x-axis at Q, then  3PQ  is equal to………
...


 2  
If the term without x in the expansion of  x 3  3  is 7315, then  is equal to………
...
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B

2
...


C

4
...


A

6
...


C

8
...


B

10
...


B

12
...


B

14
...


A

16
...


C

18
...


D

20
...


2

2
...


44

4
...


50

6
...


288

8
...


132

10
...


C

2
...


B

4
...


B

6
...


D

8
...


C

10
...


D

12
...


C

14
...


B

16
...


A

18

A OR D

19
...


B

SECTION - B
1
...


1006

3
...


372

5
...


3

7
...


14

9
...


4

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-21

PART – C (MATHEMATICS)
SECTION - A
1
...


C

3
...


B

5
...


D

7
...


C

9
...


D

11
...


D

13
...


B

15
...


A

17
...


DROP

20
...


13

2
...


105

4
...


10

6
...


16

8
...


39

10
...
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To measure the potential difference between two points, voltmeter is used
...
Now to
measure the potential difference across 600, voltmeter of 4000  is much better
than 1000  voltmeter
...


Using mirror formula
1 1 1
 
v u f
uf
v
uf
For object O1, u1 = - 15cm, f = - 20 cm, v 1 = ?
 15  20   300 cm  60cm
uf
v1  1 
u1  f  15    20 
5
For object O2, u2 = - 25cm, f = - 20 cm, v 2 = ?
 25  20   500 cm  100cm
u f
v2  2 
u2  f  25    20 
5
Hence, the distance between images formed by the mirror is d  160cm

Sol3
...


Sol4
...
(1)
MBR A
2

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-23

rA 1
 …………
...


gA MA R2A
1 1
3

  9 
2
gB
4 3
4
MBRB

Let the mass is represented as
M  Gx h y c z
x

y

M  G h c 

z

x

y

z

M  M1L3 T 2  ML2 T 1  LT 1 

 
 

M1L0 T0   M x  y  L3x  2y  z   T 2x  y  z 

 



So, on comparing we get
 x  y  1 ……………………(1)

3x  2y  z  0 ……………… (2)
2x  y  z  0 ………………
...


As we know that,
stress
FL
Y
strain
AL
g
Given: Y  2  1011,N / m2L  6mgP  , A  3mm2
4
10
Hence, M  4kgF  mgP  4 
 10N
4
10  6
2  1011 
3  10 6  L
 L  0
...


Zener diode act as a voltage regulator & it is used in reverse bias
...


Sol8
...

Therefore, assertion is false, R is true
...


  ratio of modulation index
Am  x,A c  y
A m  x,A c  2y ------------------------ (i)

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-24

Am x
 …………………… (ii)
Ac
y

1 

Am
x

Ac
2y
On solving both eqn, we get
1 2

2 1
2 

Sol10
...

Thus, the resistance of (n-1) side,
n  1 R
R2 
n
As the two parts are in parallel, So,
 R   n  1
 n   n R
R1  R2

Req 
  
R1  R2  R   n  1 
 n    n R
  

Req 

n  1 R2 

So, Req 

2

n

n
R  nR  R

n  1 R
n2

 1
1
Sol11
...
6  Z2 

 n2 n2
2
 1
Z  4 , n1  2 , n 2  4






1 1 
E  13
...
6   4  

 16 
E  40
...
Given: R 

B

0I 4  107  3

 3 T

4R
4
10

Sol13
...
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...
From ideal gas equation
PV  nRT
 volume is constant
PT
It is clear from graph that for all the gases lines of grpahs meet at the same value
...
e, temperature axis, P is zero but temperature is negative and it will be equal to
0 K or 273 C
...
Using photoelectric equation
hf  hf0  eV0
According to the question,
h  2f0   h  f0   eV1
h  2f0  f0   eV1 ………………………
...
(ii)

Using eq (i) & eq (ii), we get
h  5f0  f0   eV2
4hf0  eV2
4hf0 eV2

hf0
eV1
V2
4
V1

Or,
As we know that
KEmax  eV 

1
2
mvmax
2

 Vmax  V
Thus,

v2

v1

V2
 4 2
V1

v1 1

v2 2

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-26

Sol16
...
5  1  0
...
5  1 N-sec
For fig (c)  J 

1
 base  height
2
1
  1  0
...
375 N-sec
2

1
 base  height
2
1
  2  0
...
5 N-sec
2
So, impulse is highest for fig (b) whose area under F-t curve is maximum
...
As we know, time period of simple pendulum is
L
T  2
g
2
4
or, T2  g L
or, T 2  L
Thus, the graph between T2 & L is a straight line
...
Given: m  10 kg,  s  0
...
(i)

Along horizontal direction

F sin 30  N  mg

 N  mg  F sin30 ……
...
25  10  10
3
1
 0
...
2 N

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-27

1
3
When T2   T2  x  i
...
Given:  

' 

1
6

Temperature of hot reservoir  T1   99  C
 99  273  372 K

As we know,
T
1
  1  2  …………………(i)
T1 3
'  1 

 T2  x  
T1

1
…………
...
At the highest point vertical component of velocity is zero
...
At highest point horizontal component of velocity is not zero but vertical component of
velocity is equal to zero and because of this K
...
will not be equal to zero
...
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...


20V

10V

10

I1


1

5

I2

I3

I’

2

10

Sol2
...


R

Area  A   70cm2  70  104 m2 , B  0
...
4  70  104 sin30
7
60
 44 volt

Sol4
...

1
1
So, Total energy is equal to m 2 A 2  kA 2
2
2
1 2
 kA  0
...


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JEE-MAIN-2023 (1st February-Second Shift)-PCM-29

k

0
...
25  2



10  102



2

 50 N/m

PART – B (CHEMISTRY)
SECTION – A
Sol1
...


Sol2
...
NH3

R

CH

HX

COOH

X

NH2
 Amino acid
- amino acids exists as zwitter ion in aqueous solution as,
R CH COOH
R CH COO
NH2
Sol3
...


Sol4
...
HCl

H

NaCN , H2O
A

CN

HO

NH2

HO
LiAlH4

H

H

A

B

H2O / HCl
HO

COOH
H

B
Sol5
...
2H2O) is used to make fireproof wall boards
...
2H2O 
 CaSO4
...
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HF

+

Sol7
...

NH2

SO 3H
Sulphanilic acid
Here, both N & S are present so gives red colour in Lassigne’s test for extra element
detection
...

OCOCH3

OH
COOH

COOH
(CH3CO)2O / H+
OH

OH

COOCH3

COOH
CH3OH / H+

X is (CH3 CO)2 O / H & Y is CH3 OH / H , 
Sol9
...


Sol10
...

Sol11
...


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JEE-MAIN-2023 (1st February-Second Shift)-PCM-31

Sol12
...


rate

[(C6H5)3C-Cl]

Sol13
...
In manufacturing of urea, NH3 and CO2 are used
...
So manufacturing of urea is least responsible for global warming
...
505 -------------------------------- (ii)
Comparing (i) & (ii);
1
3
n
logk  2
...
log

Sol16
...


Sol17
...
For given equilibrium, addition of He gas at constant volume will not affect equilibrium while at
constant pressure, equilibrium shifts in forward direction & more Cl2 & PCl3 will from
...
KOH solution being basic in nature absorbs atmospheric CO2 so its concentration get
changed, therefore concentration should be checked before use in titration
...
OO bond length  H2O2 > O2F2
OH bond in H2O2 < OF bond in O2F2
...
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10% by volume means 10mL Br2 in 100 mL solution
Volume of Br2 = 10 mL
Volume of CCl4 = 90 mL
Mass of Br2 = 103
...
6g
= 144g
32
Moles of Br2 
mol
...
389 m
= 138
...


Heat released in bomb calorimeter = 200
...
3
Moles of ethane =
 0
...
01
7
C2H6 (g)  O 2 (g)  2CO2 (g)  3H2O( )
2
ng  2
...
5   8
...
225KJ
Sol3
...


i 1
n 1
i 1
0
...
2
Tf  ik f
...
372o C  372  10 3o C
Tf  1
...
86 

Sol5
...
FeS
...


Chloroliazepoxide, veronal and valium are tranquilizers while salvarsan is antibiotic
...
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OH
CH3
CH3

O
* marks are chiral carbons
...


 Ag   10 5 M
NO3   105 M

K sp
4
...
9  10 8 M
5

10
 Ag 

Using; m 
1000  M
For Ag ;


1000  105
5
  6  10 Sm 1
For NO3 ;
6  10 3 


1000  10 5
  7  105 Sm1

7  103 

For Br 

1000  4
...
2  10 8 Sm 1
Conductivity of solution   6000  7000  39
...
2  10 8 Sm 1

Sol9
...
M

 5  7 B
...
M
 5
...
M
Sol10
...
5 

3

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-34

3

dfcc
 2
...
9

PART – C (MATHEMATICS)
SECTION – A
Sol1
...

 2x  8 x   1, 2 
f ' x  
 2x  2 x   2, 3 
f '  x   0, x  4, x  1

Vertex not lies in interval
...

f  1  17, f  2   4, f  3   7
Maximum = 17, minimum = -7
max + min = 17 – 7 = 10
Sol2
...


A
12 8

14
3

 1 
f  x  f 
  1  x, f  2   ?
 1 x 
Put x  2
f  2   f  1  3 ………(i)

Put x  1
 1
f  1  f    0 ……
...
(iii)
2
Put x 

1
2

3
 1
f    f  2   ………
...
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f 1 f  2   0
f  2 f 3   0
After simplification
k   6,8 

2
1

3

k7

Sol5
...





a   2,  7, 5  , b  1, 0,1 , c  1, 2,  3 
     
 r  a  c  a, r  b  0
  
 r c a
 

 r  c  a
 
r  b  0 (given)

 
c  a  b  0









1, 2,  3     2,  7, 5    1, 0, 1  0
1  2, 2  7,  3  5  1, 0, 1  0
1  2  0  3  5   0
2

7

2
r  1, 2,  3    2,  7, 5 
7
2
2
  11
11  
11
 11   11 
r   , 0,
2
, r     
 
7 
7
7
7   7 

Sol7
...




 



AR1B if A  BC  B  A C  
C

C

 A  B   and B  A  

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-36

 A B
 R1 is reflexive and symmetric
...


Again

AR 2B if A  BC  B  A C
 V B  V  A
B A
 R2 is also reflexive, symmetric and transitive
R2 is equivalence relation
...


x1  9
x1  9, x2  9  d, x 3  9  2d, x 4  9  3d, x5  9  4d
x 6  9  5d, x7  9  6d
 d  2d  3d  4d  5d  6d
x
 3d
7
1
16  02  12
...
D = 4

x6



0

2

 12
...
3x 2  4y 2  36
dy 3x 3x 0


dx 4y 4y 0

Point  x 0 , y0  on curve
...

 2 2

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-37

2  y 0  x 0   9

Sol11
...







2n  n  1   n  4 
2n

n 1



1  1
3  1 
1
8 




  

2 n 1  2n  2 2n  1  2  n 1 2n  1 2n 



1
1  e  e1 3 e  e


2 2
2


Using e x  1 
Simplify 
Sol13
...

1 2

13  e 5

4
4
4e

 p   p  q   nq
n  p   np  q   nq
n p  q  q
 np  nq

Sol14
...
 2,1 3,4  3,0  3,6  4,5  4,6 ,  5,6 
n  A   15
n B   9
nC  9



n   A  C   B  C    6
n  A  B  C

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-38

1  ai
 1, 1  ai  b  i
bi

Sol15
...

a

 /4

Sol16
...
2x 2 y  1  xy 2 dx
dy 1  xy 2

dx
x2
Let p  y 2
2y

dp 1  px

dx
x2
dp p 1
 
dx x x
I
...
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For   2,   0
Sol19
...


5,  1,  3   1,3,5 
1,3,5
5  3  15
1  9  25



13
35

 2x  3y  z  2    x  2y  3z  6   0
x  2     y  3  2   z  1  3   6  2  0
x  2     y  3  2   z  1  3   6  2  0

…………(i)

 to 2x  y  z  1  0
2  2     1 3  2   1 1  3   0
  8
Put in (i)
6x  13y  252  46  0

 distance  7,1,1
d

6  7  13  1  25 1  46
2

2

6  13  25

d2 

2

50



830

250
83

SECTION – B


Sol1
...

a

a



2I 

 1  cos x cos 3x  cos

2



x  cos 3 x cos 3x dx

0

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-40



 /2



2I   dx   cos x cos3x  2 
0



0

 /2

0



cos2 xdx   cos3 x cos 3x dx
2







2I    0  2  cos 2 x dx   4cos6 x  3 cos 4 x dx
0

0





 2 1  
2I    2 
   4  cos6 xdx  3  cos 4 xdx
2
2

0
0
 /2

2I   


 4  2  cos6 xdx  3  2
2
6

/ 2
4

 cos xdx
0

3
 6 1 6  3 1  
 4 1 4  3 
2I 
 8

    6

2
4
2 2
4  2 
 6
 4
3 5 9
2I 


2
4
8
13 
I
16
k  13

Sol2
...


x  y  z  21
x  1  y  3  z  4  13
      13


Sol4
...
4 

5
 10
2 3

(iv) Using 3 digit 4,5,9 (4 5 9 444) =………
...
4 =
4,5,9999…………4 =

5
 20
3

5
5
4

5
5
4

4,5,9,4,5,9
...
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 2k  3 4k  6 3k  1 
C
,
,
k  1 
 k 1 k 1
8x  y  2z  8

B  2,4, 3 

A  3, 6,1 C

 2k  3   4k  6 
 3k  1 
8

 2


8
 k 1   k 1 
 k 1 
k2
 1 2 5 
C , , 
3 3 3 

 1 2 5 
C , , 
3 3 3 

x  1 y  4 z   2 
1 



1
2
3
1 
2 
5

d
...
r of  1  1, 2, 3

O

   1, 2  4, 3  2 

1
14 


 3  1 
1   1    2  2 
 3

6
3
3 


 3 

11
14
 5 130 85 
 42 , 42 , 42    1,  2 6, 17 


 a  b  c  1  26  17  10


Sol6
...


 y  2

2

 8  x  1

L f  4a cos ec 2 

m  1, cot   1

 1, 2 

8  2   16

Sol8
...
P
T11  m C1, T31  m C2 , T51  m C3
2m

C2  m C1  m C3

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JEE-MAIN-2023 (1st February-Second Shift)-PCM-42

m7
7



C5 10  3

x

10  3   3
x



x

2
2

3 x 2  21

9

10  t   t  9
10t  t 2  9
t 2  10t  9  0
t 2  9t  t  9  0
t  t  9   1 t  9   0

 t  1 t  9   0
t  1, t  9
2

3 x  1, 3x  9  x  0, x  2   0  2   4

Sol9
...
  1  9 
 2/3  
 x  x3 


Tr 1 

22



2


8 
13
 2 3 
 
3
3


2

13
 39
3

13
 39
3

22

Cr x

44  2r
 3r
3

44  2r
 3r  0
3
r4
T4 1  T5  22 C4  4  7315
4  1
  1

 1

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Title: JEE MAINS SHIFT @ FEBRUARY
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